Mistake in calculations related to vectors and norms The 2019 Stack Overflow Developer Survey...
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Mistake in calculations related to vectors and norms
The 2019 Stack Overflow Developer Survey Results Are InIs this approach for testing orthogonality/parallelity of vectors wrong as I think?Absolute value of vector not equal to magnitude of vectorEquivalent norms without Cauchy-Schwarz inequalityFind all algebraic vectorsMinimizing vector normVectors or vector fields? (Notation, physics example)Projection of vector onto another vector alternate equationA little puzzle with the way to compute the formula for Surface Integral of a Vectors FieldHow is the (global) $L^2$ norm defined for a discrete vector field?Sum of multiple vectors that equals a specific vector but their sum of norms is equal another number
$begingroup$
I am rather new to calculus, and am trying to resolve the following question. I have come to an answer, but it is not listed amongst the possible answers, so I would love to know where my reasoning failed...
Vector $mathbf u = (1,0,1)$ is split into two perpendicular vectors where one is in the direction of vector $mathbf b = (1,1,2)$. Calculate the value of $leftlVert mathbf u rightrVert ^2$$leftlVert mathbf b rightrVert ^2$$leftlVert mathbf v rightrVert ^2$$leftlVert mathbf w rightrVert ^2$.
$leftlVert mathbf b rightrVert=sqrt6$
$leftlVert mathbf u rightrVert=sqrt2$
$cosθ = frac {u·b}{leftlVert mathbf u rightrVert leftlVert mathbf b rightrVert }$=$frac{(1,0,1)(1,1,2)}{sqrt12}$=$frac {sqrt 3}{2}$, where θ refers to the angle between $mathbf u$ and $mathbf b$; therefore θ = 30 degrees.
Therefore $leftlVert mathbf v rightrVert$ is equal to $sqrt 2 · cos60^circ$=$frac {1}{2}sqrt2$.
Therefore $leftlVert mathbf w rightrVert$ is equal to $sqrt 2 · sin60^circ$=$frac {sqrt6}{2}$.
However, that leads to a result of 9 when put the norms into the equation mentioned above, and the options given are 1,2,3 and 4.
Can anybody please let me know where I went wrong?
Thank you!
calculus vectors norm
asked Mar 21 at 21:53
daltadalta
1508
$endgroup$
add a comment |
$begingroup$
I am rather new to calculus, and am trying to resolve the following question. I have come to an answer, but it is not listed amongst the possible answers, so I would love to know where my reasoning failed...
Vector $mathbf u = (1,0,1)$ is split into two perpendicular vectors where one is in the direction of vector $mathbf b = (1,1,2)$. Calculate the value of $leftlVert mathbf u rightrVert ^2$$leftlVert mathbf b rightrVert ^2$$leftlVert mathbf v rightrVert ^2$$leftlVert mathbf w rightrVert ^2$.
$leftlVert mathbf b rightrVert=sqrt6$
$leftlVert mathbf u rightrVert=sqrt2$
$cosθ = frac {u·b}{leftlVert mathbf u rightrVert leftlVert mathbf b rightrVert }$=$frac{(1,0,1)(1,1,2)}{sqrt12}$=$frac {sqrt 3}{2}$, where θ refers to the angle between $mathbf u$ and $mathbf b$; therefore θ = 30 degrees.
Therefore $leftlVert mathbf v rightrVert$ is equal to $sqrt 2 · cos60^circ$=$frac {1}{2}sqrt2$.
Therefore $leftlVert mathbf w rightrVert$ is equal to $sqrt 2 · sin60^circ$=$frac {sqrt6}{2}$.
However, that leads to a result of 9 when put the norms into the equation mentioned above, and the options given are 1,2,3 and 4.
Can anybody please let me know where I went wrong?
Thank you!
calculus vectors norm
asked Mar 21 at 21:53
daltadalta
1508
$endgroup$
1
$begingroup$
Your solution looks correct to me. Double-check the problem itself. If, for instance, those norms weren’t squared, then the result would be $3$ instead.
$endgroup$
– amd
Mar 21 at 22:43
$begingroup$
Your solution also looks correct to me, with your result of $9$ matching what I got.
$endgroup$
– John Omielan
Mar 21 at 23:08
add a comment |
$begingroup$
I am rather new to calculus, and am trying to resolve the following question. I have come to an answer, but it is not listed amongst the possible answers, so I would love to know where my reasoning failed...
Vector $mathbf u = (1,0,1)$ is split into two perpendicular vectors where one is in the direction of vector $mathbf b = (1,1,2)$. Calculate the value of $leftlVert mathbf u rightrVert ^2$$leftlVert mathbf b rightrVert ^2$$leftlVert mathbf v rightrVert ^2$$leftlVert mathbf w rightrVert ^2$.
$leftlVert mathbf b rightrVert=sqrt6$
$leftlVert mathbf u rightrVert=sqrt2$
$cosθ = frac {u·b}{leftlVert mathbf u rightrVert leftlVert mathbf b rightrVert }$=$frac{(1,0,1)(1,1,2)}{sqrt12}$=$frac {sqrt 3}{2}$, where θ refers to the angle between $mathbf u$ and $mathbf b$; therefore θ = 30 degrees.
Therefore $leftlVert mathbf v rightrVert$ is equal to $sqrt 2 · cos60^circ$=$frac {1}{2}sqrt2$.
Therefore $leftlVert mathbf w rightrVert$ is equal to $sqrt 2 · sin60^circ$=$frac {sqrt6}{2}$.
However, that leads to a result of 9 when put the norms into the equation mentioned above, and the options given are 1,2,3 and 4.
Can anybody please let me know where I went wrong?
Thank you!
calculus vectors norm
asked Mar 21 at 21:53
daltadalta
1508
$endgroup$
I am rather new to calculus, and am trying to resolve the following question. I have come to an answer, but it is not listed amongst the possible answers, so I would love to know where my reasoning failed...
Vector $mathbf u = (1,0,1)$ is split into two perpendicular vectors where one is in the direction of vector $mathbf b = (1,1,2)$. Calculate the value of $leftlVert mathbf u rightrVert ^2$$leftlVert mathbf b rightrVert ^2$$leftlVert mathbf v rightrVert ^2$$leftlVert mathbf w rightrVert ^2$.
$leftlVert mathbf b rightrVert=sqrt6$
$leftlVert mathbf u rightrVert=sqrt2$
$cosθ = frac {u·b}{leftlVert mathbf u rightrVert leftlVert mathbf b rightrVert }$=$frac{(1,0,1)(1,1,2)}{sqrt12}$=$frac {sqrt 3}{2}$, where θ refers to the angle between $mathbf u$ and $mathbf b$; therefore θ = 30 degrees.
Therefore $leftlVert mathbf v rightrVert$ is equal to $sqrt 2 · cos60^circ$=$frac {1}{2}sqrt2$.
Therefore $leftlVert mathbf w rightrVert$ is equal to $sqrt 2 · sin60^circ$=$frac {sqrt6}{2}$.
However, that leads to a result of 9 when put the norms into the equation mentioned above, and the options given are 1,2,3 and 4.
Can anybody please let me know where I went wrong?
Thank you!
calculus vectors norm
calculus vectors norm
asked Mar 21 at 21:53
daltadalta
1508
asked Mar 21 at 21:53
daltadalta
1508
asked Mar 21 at 21:53
daltadalta
1508
asked Mar 21 at 21:53
daltadalta
1508
asked Mar 21 at 21:53
daltadalta
1508
1508
1
$begingroup$
Your solution looks correct to me. Double-check the problem itself. If, for instance, those norms weren’t squared, then the result would be $3$ instead.
$endgroup$
– amd
Mar 21 at 22:43
$begingroup$
Your solution also looks correct to me, with your result of $9$ matching what I got.
$endgroup$
– John Omielan
Mar 21 at 23:08
add a comment |
1
$begingroup$
Your solution looks correct to me. Double-check the problem itself. If, for instance, those norms weren’t squared, then the result would be $3$ instead.
$endgroup$
– amd
Mar 21 at 22:43
$begingroup$
Your solution also looks correct to me, with your result of $9$ matching what I got.
$endgroup$
– John Omielan
Mar 21 at 23:08
1
1
$begingroup$
Your solution looks correct to me. Double-check the problem itself. If, for instance, those norms weren’t squared, then the result would be $3$ instead.
$endgroup$
– amd
Mar 21 at 22:43
$begingroup$
Your solution looks correct to me. Double-check the problem itself. If, for instance, those norms weren’t squared, then the result would be $3$ instead.
$endgroup$
– amd
Mar 21 at 22:43
$begingroup$
Your solution also looks correct to me, with your result of $9$ matching what I got.
$endgroup$
– John Omielan
Mar 21 at 23:08
$begingroup$
Your solution also looks correct to me, with your result of $9$ matching what I got.
$endgroup$
– John Omielan
Mar 21 at 23:08
add a comment |
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$begingroup$
Your solution looks correct to me. Double-check the problem itself. If, for instance, those norms weren’t squared, then the result would be $3$ instead.
$endgroup$
– amd
Mar 21 at 22:43
$begingroup$
Your solution also looks correct to me, with your result of $9$ matching what I got.
$endgroup$
– John Omielan
Mar 21 at 23:08